Question

$\underset{x\to 0}{lim}\left(\frac{(2+x)\sin (2+x)-2\sin 2}{x}\right)$ is equal to:

• A. $\sin 2$
• B. $\cos 2$
• C. $1$
• D. $2\cos 2+\sin 2$

Answer 1

The correct option is D $2\cos 2+\sin 2$

$\underset{x\to 0}{lim}\left(\frac{(2+x)\sin (2+x)-2\sin 2}{x}\right)$

Using l-hospital's rule, we get

$=\underset{x\to 0}{lim}\frac{(2+x)\cos (2+x)+\sin (2+x)-0}{1}$

$=2\cos 2+\sin 2$